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I'm building a power monitor (so I get notifications if the power is lost in one of the inputs). To monitor it safely, I used these 220/12 0.35VA transformers (http://www.hahn-trafo.com/english/pcb-transformers-bv20.php), I have them in a row one next to another. The transformers each are connected like this:

schematic

simulate this circuit – Schematic created using CircuitLab

The LED is for indication and the optocoupler is for connecting to Raspberry Pi.

The transformer is rated for 29mA and the actual load is lower (but not by much).

The transformers get quite hot (sticking a temperature probe between two transformers I get 70C). The manufacturers website says that the transformer is rated for "ta 70C/B". Am I still OK or do I need to add a fan?

EDIT: I connected both LEDs in series with total 2K series resistance. The temperature dropped to 58C.

EDIT2: Connected a 390R resistor in series with the diode (the PSU designer software showed that it would reduce the peak current), didn't help. Probably the 1.2W no-load dissipation as specified by the manufacturer adds quite a lot to the temperature. Added a fan to circuate the air (this is in a 1U case). Hopefully it will reduce the tempaerature.

What is a "normal temperature" for a transformer? I know that some bigger transformers get quite hot when working at specified load (no rectifier, just transformer and vacuum tube heater), don't know about the little ones - I do not have a lot of experience with very low power devices such as this.

EDIT3: With a fan the temperature is ~43C on the transformer that is furthest away from the fan. Is that considered good enough? The 58C with no fan looks to me too high, but should be below the 70C spec for the transformer (in case the fan fails).

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    \$\begingroup\$ It's a bit hard to credit the legitimacy of a monitoring circuit running hot - it suggests that there is a problem in the design or component choice. Your capacitor is too big - usually a problem on startup but it could be leaky or overvolted, too. Putting your visible and optocoupler LEDs in series rather than parallel should drastically lower your current consumption, provided there's a current level that is workable for both. Actual product designs would often very carefully drive the opto directly from the AC mains with dropping impedances but no transformer. \$\endgroup\$ – Chris Stratton Jul 12 '15 at 16:04
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    \$\begingroup\$ @ChrisStratton, just the transformers. The manufacturer says that unloaded the transformer dissipates 1.2W, so with load it most likely is more. I'll try putting the LEDs in series. \$\endgroup\$ – Pentium100 Jul 12 '15 at 16:08
  • \$\begingroup\$ @ChrisStratton I connected the visible and opto LEDs in series with total series resistance of 2K, there is enough current for both LEDs - will see how hot the transformers get now. The caps are new and 35V. I though about using CR droppers instead of transformers but decided against it for safety. I usually do not do stuff that's so low power (two LEDs in parallel = overload). \$\endgroup\$ – Pentium100 Jul 12 '15 at 16:36
  • \$\begingroup\$ Why not use a capacitor to limit the current at the transformer's secondary? Leave the transformer in place for safety, connect the LED's in series, the regular diode antiparallel across them. Drop the bulk capacitor. The only thing you need to do in software is to account for 50 or 60Hz flashing LED. \$\endgroup\$ – jippie Jul 12 '15 at 18:37
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Small transformers like this often run hotter than one might hope, it is a economising symptom. Operating them at 70 degC will cause long term reliability issues with nearby solder joints. A fan that is not monitored in some way is also a long term reliability issue. If you have space then also spread the transformers a bit apart if you can to help with cooling. Place holes in the PCB between the transformers to allow air convection.

I would suggest you use the lowest voltage output version to get the largest secondary current rating.

I would suggest that you use the largest primary voltage to get the least core losses at your rated mains voltage.

You waste half of the transformer secondary current rating by using a halfwave rectifier. Using the centre tapped secondary lets you get fullwave with two diodes or using a full bridge does the same with the single secondary.

I would use the 400V input transformer BV 202 0172 with the centre tapped 9V secondary. You should get about 4.4V on the capacitor with two diodes, with your indicator LED and OPTO in series you might not need much of a resistor, if the current limiting and voltage regulation with a small resistor value are a bit hard to characterise across part tolerances then the 12V secondary might be the way to go if the 6V secondary has enough voltage it would have the thickest secondary winding and current rating. You should be able to get a good OPTO signal with 2 to 20mA current Design for 10mA and you should be pretty safe with any of them.

The small value peak current limiting resistor will not have much effect on temperature in your transformer.

EDIT:

If you are stuck with that model of transformer you might be able to move some of the heat out of the transformer with a series resistor on the input side if you can prevent it saturating the core (keep the magnetising function linear) the iron losses will be reduced.

Using the fullwave bridge, as shown, and series connecting the LED and OPTO and reducing the current as much as possible will all help with reducing the copper losses.

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  • \$\begingroup\$ I bought the transformer that was available without special ordering, same for the opto. \$\endgroup\$ – Pentium100 Jul 12 '15 at 20:56
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The transformer rating of 29mA means that you can draw about 8mA maximum with the half-wave rectifier you have! You are exceeding the rating of the transformer, and by a considerable margin.

Suggest you use a full wave rectifier, connect the LEDs in series, and draw less than 18mA from the DC.

schematic

simulate this circuit – Schematic created using CircuitLab

If the opto LED might get disconnected (perhaps it goes through a connector) you could put a blue indicator LED across the opto output wires. That way both the green and blue LEDs will light up if the opto is disconnected while power is applied (as opposed to the green LED simply remaining dark). The blue LED would not affect the opto under normal conditions.

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  • \$\begingroup\$ Could you comment a bit on the issues raised by peak vs. average currents for the heating problem observed? It would seem like saturation might be the obvious effect of excessive peak current, but heating would seem to be linked to the time average current, unless resulting from forcing the transformer to operate in a more inefficient regime? \$\endgroup\$ – Chris Stratton Jul 12 '15 at 17:34
  • \$\begingroup\$ See the link I gave above for practical formulas for determining DC current capability from RMS rating. Even with a full wave rectifier the RMS current is higher because of the capacitor- the current is drawn in short pulses at the top of every half cycle. With a half wave rectifier it is worse- only every cycle. \$\endgroup\$ – Spehro Pefhany Jul 12 '15 at 17:38
  • \$\begingroup\$ Yes, some rating has been exceeded, but can you explain the specific mechanism by which drawing the current in pulses vs. at lower rate leading to a similar average results in greater heating? \$\endgroup\$ – Chris Stratton Jul 12 '15 at 17:39
  • \$\begingroup\$ Heating is a function of the time integral of i^2 * Rwinding, so higher peak currents cause more heating, for the same average output current. If the same average current is supplied in pulses of 10% duty cycle, the average heating will be 10x as much as if it was continuous. \$\endgroup\$ – Spehro Pefhany Jul 12 '15 at 17:43

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